# Determine the area of ​​the shape bounded by the curves

$$y=x^2 + 1, x=2, x=1, y=0$$

I've got exam today and I'm learning how to solve this type of task. The exam is about derivations mostly.

If you draw the curves roughly on a graph, you'll see that the required area is just the area under the curve $y=x^2+1$ between certain limits. Integrate $y=x^2+1$ between those limits.

Well in this case we want to integrate instead of taking the derivative.

Drawing the picture we get

To find the area under the curve, all we have to do is integrate - so we get

$$A=\int\limits_1^2(x^2+1)-0\,dx=\int\limits_1^2x^2+1\,dx=\frac{1}{3}x^3+x\bigg|_{x=1}^{x=2}=\frac{8}{3}+2-\frac{1}{3}-1=\frac{10}{3}$$

$$\int_1^2 x^2+1 \mathrm{d}x = \left[\frac{x^3}{3}+x\right]_1^2=\frac{10}{3}$$

• Adding some explanation about how one "sets up" the integral would likely be more helpful than doing the integration to the person who posted this Question. – hardmath Apr 28 '15 at 12:16
• Doing the integration can help him to understand by himself. – Hexacoordinate-C Apr 30 '15 at 1:34